An organic compound $A$ is oxidized using acidified $K_2Cr_2O_7$ solution to form compound $B$. Compound $B$ gives the $DNP$ test but not the silver mirror test. Then,compound $A$ is:
- A$2-$propanol
- B$1-$propanol
- Cethanal
- Dethanol
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Similar Questions
What will be the product obtained when formaldehyde is treated with conc. $NaOD$ followed by treatment with conc. $H_2SO_4$?
Match the example given in Column-$I$ with the name of the reaction in Column-$II$.
Column-$I$ (Example) Column-$II$ (Reaction) $(A)$ $CH_3COCl + H_2 \xrightarrow{Pd-C/BaSO_4} CH_3CHO$ $(1)$ Friedel-Crafts acylation $(B)$ $C_6H_5CHO \xrightarrow{NaOH} C_6H_5CH_2OH + C_6H_5COONa$ $(2)$ $HVZ$ reaction $(C)$ $C_6H_6 + CH_3COCl \xrightarrow{AlCl_3} C_6H_5COCH_3$ $(3)$ Aldol condensation $(D)$ $R-CH_2COOH \xrightarrow{Br_2/Red P} R-CH(Br)COOH$ $(4)$ Cannizzaro's reaction $(E)$ $CH_3-CN \xrightarrow{SnCl_2/HCl, H_2O/H^+} CH_3-CHO$ $(5)$ Rosenmund's reduction $(F)$ $2CH_3CHO \xrightarrow{NaOH} CH_3-CH=CH-CHO$ $(6)$ Stephen's reaction
| Column-$I$ (Example) | Column-$II$ (Reaction) |
|---|---|
| $(A)$ $CH_3COCl + H_2 \xrightarrow{Pd-C/BaSO_4} CH_3CHO$ | $(1)$ Friedel-Crafts acylation |
| $(B)$ $C_6H_5CHO \xrightarrow{NaOH} C_6H_5CH_2OH + C_6H_5COONa$ | $(2)$ $HVZ$ reaction |
| $(C)$ $C_6H_6 + CH_3COCl \xrightarrow{AlCl_3} C_6H_5COCH_3$ | $(3)$ Aldol condensation |
| $(D)$ $R-CH_2COOH \xrightarrow{Br_2/Red P} R-CH(Br)COOH$ | $(4)$ Cannizzaro's reaction |
| $(E)$ $CH_3-CN \xrightarrow{SnCl_2/HCl, H_2O/H^+} CH_3-CHO$ | $(5)$ Rosenmund's reduction |
| $(F)$ $2CH_3CHO \xrightarrow{NaOH} CH_3-CH=CH-CHO$ | $(6)$ Stephen's reaction |
Medium
View SolutionIn the scheme given below,the total number of intra molecular aldol condensation products formed from '$Y$' is
Give the reaction of $(a)$ acetaldehyde and $(b)$ acetone with $(i)$ $NaHSO_{3}$ and $(ii)$ $HCN$.
Medium
View SolutionWhich of the following compounds ($i$ through $v$) should not be classified as an acetal?
Medium
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